What is the limit as h approaches 0 of cos(h) ?

The limit as h approaches 0 of cos(h) is 1

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limit as h approaches 0 of cos(h)

h \to 0 lim (cos (h))

1

Solution steps

h \to 0 lim (cos (h))

Plug in the value

h = 0

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

x \to - 1 lim (\frac{1 - x ^{4}}{x + 1})

x \to 0 lim (sin (x) - tan (x))

x \to 0 + lim (lo g_{e} (x))

x \to 5 + lim (f (x))

x \to \infty lim (- 2 ln (x))

What is the limit as h approaches 0 of cos(h) ?
The limit as h approaches 0 of cos(h) is 1